Trigonometry & Inverse Trigonometry
Properties of Triangle - Circumradius
Grade 11
Question:
<p>94. The diameter of the circumcircle of a triangle with sides 5, 6 and 7 is</p>
<p>(a) \(\dfrac{3\sqrt{6}}{2}\)</p>
<p>(b) \(2\sqrt{6}\)</p>
<p>(c) \(\dfrac{35}{4\sqrt{6}}\)</p>
<p>(d) \(\dfrac{9}{\sqrt{2}}\)</p>
<p>(e) none of these</p>
Step-by-Step Solution
Key Concept: Use the extended sine rule (2R = a/sin A) where R is the circumradius. First find sin A using the cosine rule and area formula, then calculate the diameter 2R.
<p><strong>Step 1:</strong> Use Heron's formula to find the area. With sides a=5, b=6, c=7:<br>s = (5+6+7)/2 = 9<br>Area K = √[s(s-a)(s-b)(s-c)] = √[9·4·3·2] = √216 = 6√6</p><p><strong>Step 2:</strong> Apply the formula R = abc/(4K) for circumradius:<br>R = (5·6·7)/(4·6√6) = 210/(24√6) = 35/(4√6)</p><p><strong>Step 3:</strong> Rationalize:<br>R = 35√6/(4·6) = 35√6/24</p><p><strong>Step 4:</strong> The diameter is:<br>2R = 2 · (35√6/24) = 35√6/12</p><p>∴ Answer: C</p>
Correct Answer: C